1. **State the problem:** We need to solve for $n \in \mathbb{N}$ the equation $$ (\log_2 3)(\log_3 4)(\log_4 5) \cdots (\log_{n-1} n)(\log_n (n+1)) = 11. $$ 2. **Recall the change of base formula and telescoping property:** Using the change of base formula, $$ \log_a b = \frac{\log_c b}{\log_c a} $$ for any positive base $c \neq 1$. 3. **Simplify the product using telescoping:** Observe that $$ (\log_2 3)(\log_3 4) = \frac{\log 3}{\log 2} \times \frac{\log 4}{\log 3} = \frac{\log 4}{\log 2} = \log_2 4. $$ Similarly, $$ (\log_2 3)(\log_3 4)(\log_4 5) = \log_2 5, $$ and so on. 4. **Generalize the telescoping product:** The entire product simplifies to $$ (\log_2 3)(\log_3 4) \cdots (\log_n (n+1)) = \log_2 (n+1). $$ 5. **Set the simplified expression equal to 11:** $$ \log_2 (n+1) = 11. $$ 6. **Solve for $n$:** Rewrite in exponential form: $$ n+1 = 2^{11}. $$ Calculate: $$ n+1 = 2048 \implies n = 2047. $$ **Final answer:** $$ \boxed{2047} $$ This means the value of $n$ that satisfies the equation is 2047.